# Is there a decomposition $U U^T$?

We know that there exist the Choleski decomposition

$M = L L^T$

where $M$ is a positive definite matrix and $L$ a lower triangular one. Does it exist a similar decomposition in

$M = U U^T$

with $U$ upper triangular?

• I think you would need $M = U^TU$ In fact this is what Matlab knows as its default Cholesky decomposition. au.mathworks.com/help/matlab/ref/chol.html Sep 14, 2015 at 12:19
• Let $W_0$ be the permutation matrix of the permutation which sends every $k \in \left\{1,2,\ldots,n\right\}$ to $n+1-k$. (Here, $n$ is the size of our $M$.) Then, the matrix $W_0 M W_0^T$ is symmetric and positive definite. Thus, let $W_0 M W_0^T = L L^T$ be the Cholesky decomposition of $W_0 M W_0^T$. Set $U = W_0 L W_0^T$. Show that $U$ is upper-triangular, and show that $M = U U^T$. (Note that I am speaking of $W_0^T$ for aesthetical reasons only; of course, $W_0^T = W_0$.) Sep 14, 2015 at 14:16
• Here is a more conceptual way to rewrite this proof: The map $\operatorname{M}_n\left(\mathbb{R}\right) \to \operatorname{M}_n\left(\mathbb{R}\right),\ A \mapsto W_0 A W_0^T$ is a ring automorphism which commutes with taking transposes and preserves the properties of being symmetric and of being positive definite; however, it interchanges the property of being lower triangular with the property of being upper triangular. So the $M = LL^T$ decomposition and the $M = UU^T$ decomposition are equivalent (although, of course, not for a single fixed matrix $M$). Sep 14, 2015 at 14:20